Question

functions over the interval $0 \leq x \leq 1$?
the exponential grows at half the rate of the quadratic.
the exponential grows at the same rate as the quadratic.
the exponential grows at twice the rate of the quadratic.
the exponential grows at four times the rate of the quadratic.

Explanation:

Step1: Identify the functions

The quadratic function (parabola) passes through \((0,0)\) and \((1,1)\) (assuming vertex at origin, so \(y = x^2\)). The exponential function passes through \((0,1)\) and \((1,2)\) (so \(y = 2^x\) or similar, but let's check rates).

Step2: Calculate rate of change (slope)

For quadratic (\(y = x^2\)) on \([0,1]\): rate = \(\frac{f(1)-f(0)}{1 - 0}=\frac{1 - 0}{1}=1\).

For exponential (let's say \(y = 2^x\)) on \([0,1]\): rate = \(\frac{f(1)-f(0)}{1 - 0}=\frac{2 - 1}{1}=1\)? Wait, no, maybe the exponential is \(y = e^x\)? Wait, no, the point at \(x=1\) for exponential is 2, quadratic is 1. Wait, initial point: at \(x=0\), quadratic is 0, exponential is 1. At \(x=1\), quadratic is 1, exponential is 2. So rate of quadratic: \(\frac{1 - 0}{1 - 0}=1\). Rate of exponential: \(\frac{2 - 1}{1 - 0}=1\)? Wait, no, that's same? Wait, but maybe the quadratic is \(y = x^2\), exponential is \(y = 1 + x\)? No, the graph: the exponential starts at (0,1), goes to (1,2); quadratic starts at (0,0), goes to (1,1). So the rate (average rate) for quadratic is \(\frac{1 - 0}{1 - 0}=1\), for exponential is \(\frac{2 - 1}{1 - 0}=1\). Wait, but maybe the options: "The exponential grows at the same rate as the quadratic" – but wait, no, maybe I misread. Wait, the quadratic is the parabola (opening up, vertex at origin), exponential is the curve starting at (0,1), increasing. At \(x=0\), quadratic: 0, exponential: 1. At \(x=1\), quadratic: 1, exponential: 2. So the change in y for quadratic: 1 - 0 = 1. For exponential: 2 - 1 = 1. So average rate over [0,1] is same. Wait, but maybe the question is about average rate. So the rates are equal.

Wait, but let's recheck. Quadratic function: let's assume it's \(y = x^2\) (since it passes through (0,0) and (1,1)). Exponential function: passes through (0,1) and (1,2), so \(y = 2^x\) (since \(2^0 = 1\), \(2^1 = 2\)). The average rate of change for \(y = x^2\) on [0,1] is \(\frac{1^2 - 0^2}{1 - 0}=1\). For \(y = 2^x\) on [0,1] is \(\frac{2^1 - 2^0}{1 - 0}=1\). So their average rates are equal. Wait, but maybe the options: "The exponential grows at the same rate as the quadratic" – but wait, no, maybe the quadratic is \(y = x^2\), exponential is \(y = e^x\)? No, \(e^1 \approx 2.718\), but the point is 2. So maybe the exponential is \(y = 2^x\), quadratic \(y = x^2\). So rates are same. Wait, but the options: "The exponential grows at the same rate as the quadratic" – is that correct? Wait, maybe I made a mistake. Wait, the quadratic's rate (average) is 1, exponential's rate is 1. So yes, same rate.

Answer:

The exponential grows at the same rate as the quadratic. (Assuming the options are as given, and the average rate of change over [0,1] is same for both.)